高等结构动力学大作业

Advanced Structural Dynamics Project

The dynamic response and stability analysis of the beam under vertical excitation

Instructor：Dr. Li Wei

Name：

Student ID：

1.Problem description and the purpose of the project

1.1 calculation model

An Eular beam subjected to an axial force. Please build the differential equation of motion and use a proper difference method to solve this differential equation. Study the dynamic stability of the beam related to the frequency and amplitude of the force. As shown in the Fig 1.1.

Fig1.1

1.2 purpose and process arrangement

a.learning how to create mathematical model of the continuous

system and select proper calculation method to solve it.

b.learning how to build beam vibration equation and solve Mathieu

equation.

ing Floquet theory to judge vibration system’s stability and

analyze the relationship among the frequency and amplitude of the force and dynamic response.

This project will introduce the establishment of the mathematical model of the continuous system in section 2, the movement equation and the numerical solution of using MATLAB in section 3, Applying Floquent theory to study the dynamic stability of the beam related to the frequency and amplitude of the force in section 4. In the last of the project, we get some conclusions in section 5.

2. the mathematical model of the system

The geometric model of the beam and force-simplified diagram is shown in Fig.2.1.We assume that its stiffness(EI) is constant and the deflection of the beam is small, and the boundary conditions is simply support. Now the beam subjected to an axial force. We assume the force is equal to 0cos P t ω. F=f 0coswt y

x

Fig.2.1

We select the length of x ∆ in any position of the beam, the free-body diagram is shown in Fig.2.2.

Fig.2.2

Using equations of movement equilibrium, that is to say:

+↑()y y F m a ∑=∆ （1）

0G M +∑= （2） From equation (1), we will get:

22),(),(t

y x A t x x S t x S ∂∂∆=∆+-ρ （3）

Divide equation (3) by x ∆ and take the limit:

22x

y A x S ∂∂=∂∂-ρ （4） Then synthesize equation (2),we can get:

0),()],(),([),(),(=∆∆+--∆++-∆+x t x x S t x y t x x y F t x M t x x M （5）

Divide equation (5) by x ∆ and take the limit:

S x

y F x M =∂∂+∂∂ （6） Combine equation (4) with (6):

0222222=∂∂+∂∂+∂∂t

y A x y F x M ρ （7） And 22),(x

y EI t x M ∂∂= （8） Combine equation (7) with (8):

0)(22222222=∂∂+∂∂+∂∂∂∂t

y A x y F x y EI x ρ （9） We know EI is a constant, so

0)(222244=∂∂+∂∂+∂∂t

y m x y t F x y EI （10） In equation (10), m is the mass of unit length. Now we will use assumed-modes method. Named l

x n t T t x y n πsin )(),(=,so: 0sin )(22244422=⎥⎦

⎤⎢⎣⎡-+l x n T l n t F T l n EI dt T d m n n n πππ （11） 0))(1(222=-+n n

on n T F t F P dt T d n=1,2,...... （12） In the equation (12)

2

2222

2,l EI n F m EI l n P n on ππ==